Given a functional $S=S\{Y[X(r)]\}$, is the following "chain rule" valid?

$$\frac{\delta S\{Y[X]\}}{\delta X(r)}=\frac{\partial Y(r)}{\partial X(r)}\frac{\delta S[Y]}{\delta Y(r)}$$


2 Answers 2


The chain rule for functional differentiation is just the continuum generalisation of the usual chain rule for differentiation of a function of many variables $f(y_1,y_2,\ldots,y_N) = f(\mathbf{y})$, which reads $$ \frac{\partial f(\mathbf{y})}{\partial x_i(\mathbf{y})} = \sum\limits_{j=1}^N\frac{\partial y_j}{\partial x_i}\frac{\partial f}{\partial y_j}. $$ The continuum limit amounts to sending the number of variables $N\to\infty$, and defining a new continuous index $r$ such that $j\to rN$. Then you just change your notation to agree with the continuous nature of the new index $r$, e.g. $x_j \to X(r)$, $f(\mathbf{x}) = f(\{x_j\}) \to F[X(r)]$ etc. In our example, this means substituting the above sum over the discrete index $j$ by an integral over a continuous index, finding the chain rule for functional differentiation: $$\frac{\delta F[Y]}{\delta X(r)} = \int \mathrm{d}s\,\frac{\delta Y(s)}{\delta X(r)}\frac{\delta F[Y]}{\delta Y(s)}.$$ You can get any functional calculus identity you want along the same lines. Just think about what happens in ordinary multivariate calculus with a finite number of variables. Then take the number of variables to infinity.

  • $\begingroup$ Sorry to accept it so late. Thank you for your effort! $\endgroup$
    – Machine
    Dec 2, 2012 at 7:16
  • $\begingroup$ It is very difficult to find approachable resources on functional calculus beyond the basics, so this answer was incredibly helpful! There is one thing I still can't wrap my head around, though: what is the domain of integration of this integral? Is it from minus infinity to infinity? $\endgroup$
    – Godzilla
    Jun 8, 2020 at 17:03
  • $\begingroup$ it would be the domain of the functional. Ex: if the functional was $\int_{0}^{1} (f+f')$ then this domain of integration would be from $0$ to $1$. Note most functionals, that is functions which take functions as inputs and produce as output complex numbers, Are representable as an integral of a (function of functions) over some complex domain. In the case above the functional was integral 0 to 1 (f + f'). the Function of functions was just f -> f+f', and the domain was the interval from 0 to 1. $\endgroup$ Jan 3, 2021 at 4:39
  • 1
    $\begingroup$ what does $\frac{\delta Y(s)}{\delta X(r)}$ mean? $\endgroup$
    – wlad
    Jan 6, 2021 at 14:46
  • $\begingroup$ An example would help: what is $\frac{\delta F[Y[f](r)]}{\delta f(r)}$ where $Y[f](r) = \int_0^r X(t)\, dt$? $\endgroup$
    – wlad
    Jan 6, 2021 at 15:45

Yes. There is a chain rule for functional derivatives. But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot.

Consider the scalar function $f: (y_1,...,y_n) \mapsto g(x_1(\mathbf{y}),...,x_n(\mathbf{y}))$, which is the composition of $g$ with the coordinate functions $x_i(\mathbf{y})$.

The partial derivative of $f$ with respect to $y_j$ at the point $\mathbf{y}$ is given by the chain rule

$$ \frac{\partial f (\mathbf{y}) }{\partial y_j}= \sum_{i = 1}^N \frac{\partial g(\mathbf{x}(\mathbf{y}))}{\partial x_i} \frac{\partial x_i(\mathbf{y})}{\partial y_j}. $$

Now, the continuum generalization of the coordinate functions $x_i$ is the one-parameter family of functionals

$$ \lambda(s) = \Lambda[f;s], \quad s \in \mathbb{R}, $$

where $\Lambda[f;s]$ means that $\Lambda$ is a functional of $f$ and a function of $s$.

A functional $G$ can only act on $\Lambda$ considered as a function of the parameter $s$. Therefore, $G[\Lambda[f;s]]$ is meant to be understood as $G[\lambda]$. Contrary to the claims of many physics textbooks, there is no such thing as the functional of a functional.

Since a variation of $f$ will change the value of $\lambda$, ultimately it results in a variation of $G$. Therefore, $G[\Lambda[f;s]]=G[\lambda]$ can be considered as a functional $F[f]$. The partial functional derivative of $F$ with respect to the local change of $f$ at $x$ is given by [1] Appendix A:

$$ \frac{\delta F [f]}{\delta f (x)} = \int \mathrm{d} y \frac{\delta G [\lambda]}{\delta \lambda (y)} \frac{\delta \Lambda [f ; y]}{\delta f (x)} . $$

This is the generalization of the chain rule for functionals.

[1] E. Engel, R.M. Dreizler, Density Functional Theory: An Advanced Course, Springer Science & Business Media, 2011.


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